Number Sequence HDU

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one. 

InputThe first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000]. 
OutputFor each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead. 
Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1
代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
int n, m;
int a[maxn];
int b[maxn];
int nex[maxn];

void getnext()
{
    int i = 1, j = 0;
    nex[1] = 0;
    while (i < m)
    {
        if (j == 0 || b[i] == b[j])
        {
            i++, j++;
            if (b[i] != b[j]) nex[i] = j;
            else nex[i] = nex[j];
        }
        else j = nex[j];
    }
}

int kmp()
{
    int i, j; i = j = 1;
    while (i <= n && j <= m)
    {
        if (j == 0 || a[i] == b[j]) i++, j++;
        else j = nex[j];
    }
    if (j > m)
        return i - m;
    else return -1;
}

int main()
{
    int T; cin >> T;
    while (T--)
    {
        cin >> n >> m;
        for (int i = 1; i <= n; i++)
            scanf("%d", a + i);
        for (int i = 1; i <= m; i++)
            scanf("%d", b + i);
        getnext();
        int res = kmp();
        cout << res << endl;
    }
}